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What adjustable linear regulator to use?

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bd139:
Good point. I've never run my LT3080 board without a load.

LukeW:
The LT3080 does seem a little expensive (well duh it's LT) but it seems really simple and easy, quality regulation - better than setting up extra parts and a negative rail converter on an LM317 to make it go down to zero, which will be nearly as expensive and not as good.

Here's a challenge.

The program current on the LT3080 is 10 uA.
Therefore, to go from 0-5V you would have a 500k \$\Omega\$ variable resistor.

But what if the pot was already committed, already glued to the devices, absolutely can't be changed and it's a 10k \$\Omega\$ pot?

Is there any nice way that we could set 0-5V using a pot which must be 10k?

soldar:

--- Quote from: LukeW on March 18, 2019, 02:04:56 am --- The program current on the LT3080 is 10 uA.
Therefore, to go from 0-5V you would have a 500k \$\Omega\$ variable resistor.

But what if the pot was already committed, already glued to the devices, absolutely can't be changed and it's a 10k \$\Omega\$ pot?

Is there any nice way that we could set 0-5V using a pot which must be 10k?
--- End quote ---

The answer to your question is right there in the application notes.

Bassman59:
LT3080 is pretty great. Just mind the Vc control voltage, which has to be something like 1.5 V above the output. So unless you have a separate higher-voltage supply it can’t really be used as an LDO, even though Vin to Vout min is 300 mV or thereabouts. (Numbers from memory, I don’t have the data sheet right here.)

LukeW:
Thanks for pointing out that datasheet schematic I overlooked.

R1 and R2 need to be quite small in order to get as close to 0 as possible when Rset is 0.
Let's say R1 is 3.3k and R2 is 68 ohms. RSet is 0-10k.

That should give us Vout = (3.3k*10uA) + (10k*10uA)*(1+(3300/68)) = 4.99 volts at the 10k pot maximum.
If I've calculated that correctly. Can you confirm that is right?

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